i usually attempt this type of question by trial and error. Usually my choice of guess is range from -3 to 3. Since, using x=3 is one of the factor, it help reduce the equation to this form; x^2 + 2x +1. you can solve the above equation using quadratic formula; x^3-x^2-5x-3 = (x-3)(x^2 + 2x + 1) = (x-3)(x-1)^2
to find a root you could use newton's method: x(n+1) = x(n) - f(x(n))/f'(x(n)) if you have a ti calculator try this: 2->x enter x - (x^3-x^2-5x-3)/(3x^2-2x-5) -> x enter (5 times) you should see it approach 3. This is a good way if you can't really think of a root but have a calculator. Then proceed by division as kenius describes.
where "->" is the "sto>" button
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